The joint probability distribution of thr random variables X and Y is given below: f(x,y)={(cxy , 0<x<2,0<y<x),(0,, \text{ other }):}

Step 1
Given:
$f(x,y)=\{\begin{array}{ll}cxy& 0<x<2,0<y<x\\ 0& \text{ other }\end{array}$
a. value of constant c,
${\displaystyle {\int }_0^2{\int }_0^{x}cxydydx=1}$
${\displaystyle {\int }_0^{2}cx{\left[\frac{y^2}{2}\right]}_0^{x}dx=1}$
${\displaystyle \frac{c}{2}{\int }_0^{2}x\times x^{2}dx=1}$
${\displaystyle \frac{c}{2}{\left[\frac{x^4}{4}\right]}_0^2=1}$
2c=1
${\displaystyle c=\frac{1}{2}}$
Step 2
b.
we have given a joint pdf,
$f(x,y)=\{\begin{array}{ll}cxy& 0<x<2,0<y<x\\ 0& \text{ other }\end{array}$
marginal pdf of x,
${\displaystyle f_1\left(x\right)={\int }_0^{x}f(x,y)dy}$
${\displaystyle f_1\left(x\right)={\int }_0^x\frac{1}{2}xydy}$
${\displaystyle f_1\left(x\right)=\frac{1}{2}x\times {\left(\frac{y^2}{2}\right)}_0^x}$
$f_1(x)=\{\begin{array}{ll}\frac{1}{4}{x}^3& 0<x<2\\ 0& \text{ otherwise }\end{array}$
${\displaystyle E\left(x\right)={\int }_0^{2}x\times f_1\left(x\right)dx}$
${\displaystyle E\left(x\right)={\int }_0^2\times \frac{x^3}{4}dx}$
${\displaystyle E\left(x\right)=\frac{1}{4}{\left(\frac{x^5}{5}\right)}_0^2}$
${\displaystyle E\left(x\right)=\frac{8}{5}}$
Step 3
${\displaystyle E\left(x^2\right)={\int }_0^2{x}^2\times f_1\left(x\right)dx}$

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